\newproblem{lay:5_1_9}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.1.9}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find a basis for the eigenspace corresponding to each of the eigenvalues of $A=\begin{pmatrix}3 & 0 \\ 2 & 1\end{pmatrix}$, $\lambda=1,3$
}{
  % Solution
	We need to find the set of vectors such that for each eigenvalue they meet
	\begin{center}
		$A\mathbf{v}=\lambda\mathbf{v} \Rightarrow (A-\lambda I)\mathbf{v}=\mathbf{0}$
	\end{center}
	\underline{$\lambda=1$}\\
	\begin{center}
		$\left(\begin{pmatrix}3 & 0 \\ 2 & 1\end{pmatrix}-\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\right)\mathbf{v}=\mathbf{0}$ \\
		$\begin{pmatrix}2 & 0 \\ 2 & 0\end{pmatrix}\mathbf{v}=\mathbf{0}$ \\
	\end{center}
	The general solution of this homogeneous equation system is $\mathbf{v}=\begin{pmatrix}0\\x_2\end{pmatrix}$. This is the eigenspace associated to the
	eigenvalue $\lambda=1$ and one of its basis is $\{(0,1)\}$.\\
	
	\underline{$\lambda=3$}\\
	\begin{center}
		$\left(\begin{pmatrix}3 & 0 \\ 2 & 1\end{pmatrix}-3\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\right)\mathbf{v}=\mathbf{0}$ \\
		$\begin{pmatrix}0 & 0 \\ 2 & -2\end{pmatrix}\mathbf{v}=\mathbf{0}$ \\
	\end{center}
	The general solution of this homogeneous equation system is $\mathbf{v}=\begin{pmatrix}x_2\\x_2\end{pmatrix}$. This is the eigenspace associated to the
	eigenvalue $\lambda=3$ and one of its basis is $\{(1,1)\}$.
}
\useproblem{lay:5_1_9}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
